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# Cofunction Identities

Trigonometry deals with the measurements of arms and angles of right-angled triangles. There are different types of functions that are associated with our good old six trigonometric ratios - sin, cot, cos, cosec, tan and sec. In this article, we are going to learn about cofunctions and cofunction identities. We will throw light on cofunction identities in degrees as well as radians, their proofs, and examples based on cofunction identities.

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## Definition

In mathematics, the cofunctions are often seen and used. Two functions are said to be cofunctions for each other when their arguments together make complementary angles. More elaborately, two functions f and g are said to be cofunctions of each other if f(P) = g(Q), provided that P and Q are complementary angles.

Cofunctions are typically defined in the reference of trigonometric functions. In trigonometry, sine and cosine are cofunctions. Likewise, secant and cosecant are cofunctions and tangent and cotangent is cofunctions of each other. Therefore, the word "co" is prefixed in cos, cot, and cosec.

Cofunction identities in radians are written as given below.

(i) sin ($\frac{\pi}{2}$ - x) = cos x

(ii) cos ($\frac{\pi}{2}$ - x) = sin x

(iii)
csc ($\frac{\pi}{2}$ - x) = sec x

(iv)
sec ($\frac{\pi}{2}$ - x) = csc x

(v)
tan ($\frac{\pi}{2}$ - x) = cot x

(vi)
cot ($\frac{\pi}{2}$ - x) = tan x

## Cofunction Identities in Degrees

Cofunction identities in degrees can be expressed as under.

(i) sin (90$^{\circ}$ - x) = cos x

(ii)
cos (90$^{\circ}$ - x) = sin x

(iii)
csc (90$^{\circ}$ - x) = sec x

(iv)
sec (90$^{\circ}$ - x) = csc x

(v)
tan (90$^{\circ}$ - x) = cot x

(vi)
cot (90$^{\circ}$ - x) = tan x

## Proofs

Proof of sin ($\frac{\pi}{2}$ - x) = cos x

Using formula sin (A - B) = sin A cos B - cos A sin B

$sin$ ($\frac{\pi}{2}$ $- x) = sin$ $\frac{\pi}{2}$ $cos x - cos$ $\frac{\pi}{2}$  $sin x$

= 1. cos x - 0. sin x

= cos x

Proof of cos ($\frac{\pi}{2}$ - x) = sin x

Using formula cos (A - B) = cos A cos B + sin A sin B

$cos$ $($$\frac{\pi}{2} - x) = cos \frac{\pi}{2} cos x + sin \frac{\pi}{2} sin x = 0. cos x + 1. sin x = sin x Proof of csc (\frac{\pi}{2} - x) = sec x csc\ ($$\frac{\pi}{2}$ $-\ x)\ =$ $\frac{1}{sin (\frac{\pi}{2} - x)}$

= $\frac{1}{cos x}$

= sec x

Proof of sec ($\frac{\pi}{2}$ - x) = csc x

$sec\ ($$\frac{\pi}{2} -\ x)\ = \frac{1}{cos (\frac{\pi}{2} - x)} = \frac{1}{sin x} = csc x Proof of tan (\frac{\pi}{2} - x) = cot x tan\ ($$\frac{\pi}{2}$ $-\ x)$ = $\frac{sin(\frac{\pi}{2} - x)}{cos(\frac{\pi}{2} - x)}$

= $\frac{cos x}{sin x}$

= cot x

Proof of cot ($\frac{\pi}{2}$ - x) = tan x

$cot\ ($$\frac{\pi}{2}$ $-\ x)$ = $\frac{cos(\frac{\pi}{2} - x)}{sin(\frac{\pi}{2} - x)}$

= $\frac{sin x}{cos x}$

= tan x

(i) sin ($\frac{\pi}{2}$ + x) = cos x

(ii) cos ($\frac{\pi}{2}$ + x) = -sin x

(iii)
csc ($\frac{\pi}{2}$ + x) = sec x

(iv)
sec ($\frac{\pi}{2}$ + x) = -csc x

(v)
tan ($\frac{\pi}{2}$ + x) = -cot x

(vi)
cot ($\frac{\pi}{2}$ + x) = -tan x

## Examples

The examples based on cofunction identities are discussed below.

Example 1 : Evaluate the expression $sin^{2}\ 18^{\circ}\ +\ sin^{2}\ 50^{\circ}\ +\ sin^{2}\ 40^{\circ}\ +\ sin^{2}\ 72^{\circ}$.

Solution : $sin^{2}\ 18^{\circ}\ +\ sin^{2}\ 50^{\circ}\ +\ sin^{2}\ 40^{\circ}\ +\ sin^{2}\ 72^{\circ}$

= $(sin 18^{\circ})^{2}\ +\ (sin\ 50^{\circ})^{2}\ +\ (sin\ 40^{\circ})^{2}\ +\ (sin\ 72^{\circ})^{2}$

We know that 18 + 72 = 90 and 40 + 50 = 90, hence

$(sin(90\ -\ 72)^{\circ})^{2}\ +\ (sin\ (90\ -\ 40)^{\circ})^{2}\ +\ sin^{2}\ 40^{\circ}\ +\ sin^{2}\ 72^{\circ}$

= $(cos\ 72^{\circ})^{2}\ +\ (cos\ 40^{\circ})^{2}\ +\ sin^{2}\ 40^{\circ}\ +\ sin^{2}\ 72^{\circ}$

= $cos^{2}\ 72^{\circ}\ +\ cos^{2}\ 40^{\circ}\ +\ sin^{2}\ 40^{\circ}\ +\ sin^{2}\ 72^{\circ}$

= $(cos^{2}\ 72^{\circ}\ +\ sin^{2}\ 72^{\circ})\ +\ (cos^{2}\ 40^{\circ}\ +\ sin^{2}\ 40^{\circ})$

= 1 + 1 = 2
Example 2 : Find the value of angle A from the given cofunction identity, assuming that angles involved are acute angles of right triangle :

$cos\ (A\ +\ 4^{\circ})\ =\ sin\ (3A\ +\ 2^{\circ})$

Solution : $cos\ (A\ +\ 4^{\circ})\ =\ sin\ (3A\ +\ 2^{\circ})$

Since $A\ +\ 4^{\circ}$ and $3A\ +\ 2^{\circ}$ are acute angles of a right triangle, hence

$A\ +\ 4^{\circ}\ +\ 3A\ +\ 2^{\circ}\ =\ 90^{\circ}$

$4A\ +\ 6^{\circ}\ =\ 90^{\circ}$

$4A\ =\ 84^{\circ}$

A = 21$^{\circ}$
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